Problem Solving

If two of the four expressions: x+y, x+5y, x-y, 5x-y are chosen at random, what is the probability that their product will be in the form of x^2 -(by)^2, where b is an integer?

The correct answer is [spoiler]1/6[/spoiler]

Please help.

djkvakin wrote:If two of the four expressions: x+y, x+5y, x-y, 5x-y are chosen at random, what is the probability that their product will be in the form of x^2 -(by)^2, where b is an integer?

The correct answer is [spoiler]1/6[/spoiler]

Please help.

There are 4C2 or 6 ways to pull out a product pair from those four expressions. Among these product pairs, only (x+y)*(x-y) will produce a difference of squares.

To see this, reason as follows. We want y^2 to be subtracted from x^2, so in one of our multipliers we want y to be subtracted; in the other one, we want to be adding the terms (we need pos*neg to make neg). Focussing on (x + y) first, we see that we can have either (x + y) * (x - y) or (x + y) * (5x - y). (x + y) * (x - y) will certainly produce a difference of squares (that's one out of 6 so far). However, you can see that (x + y) * (5x - y) will have an "xy" term (that is, if you were to expand, the "xy" terms will not cancel out).

Now, looking at x + 5y, we can have (x + 5y) * (x - y) or (x + 5y) * (5x - y). In both of these, we will have intermediate "xy" terms.

Thus, there is only 1 product pair (out of 6) that will give us a difference of squares.

Using Testluv logic another way to solve it is 2/4 * 1/3 = 1/6

Thank you, testluv